B. -4.08 D. You must first determine r by using the equation [Fair = -12.00 D = (1.50 – 1) / r]. You determine that r = -0.0417m. Using the new equation [Fwater = (1.50 – 1.33) / -0.0417m].

A. The insect appears to be 53.20cm above the surface. To solve, first draw a diagram with the insect above the water with a line acting as the surface of the pond. Now use the equation (l’/n’) = (l/n). It should read (l’/1.33) = (40/1); l = 53.20

C. The true power is +7.47 D. To solve, first determine r using the clock reading: [Flc = +6.00 D = (1.53 – 1) / r; r = +0.08833m. Now solve for true with Flens = (1.66 – 1) / r = 0.66/0.08833 = +7.47 D)

D. Purkinje image IV is the only image that is real and inverted. The remaining three are virtual and upright.

D. The power in the 180 meridian is -1.00. On an optical cross, -2.00DS falls on the 90 meridian. Subtract the astigmatic difference of -3.00 to find the resulting -1.00DS.

A. The apparent depth is -2.0 cm. Apparent depth is determined using the index of refraction of both mediums and the given depth. Therefore, n2/l’ = n1/l → (1.0/l’ = -1.5/0.03m = -50 cm) → 1/-50 = l’ = -2.0cm.

C. The power of the thin, Crown glass lens (n = 1.523) is -13.08 D. First draw a picture of this image (should look like two “C”s). First find F1 = (n2 – n1)/r, then F2 with the same equation. Done correctly will result in F1 = +13.075 D; F2 = -26.15 D; F = F1 + F2 → F = 13.075 + (-26.15) = -13.075 D.

D. This CT pattern is indicative of against-the-rule astigmatism. WTR astigmatism is characterized by a bow tie pattern with vertical steepening. Keratoconus would present with inferior steepening. Pellucid marginal degeneration scans are known for a kissing dove or crab claw appearance.

D. The correct answer is +1.00 -3.00 x 090. Although cylinder lines are cleared at -2.00 D, the difference between both meridians is the DC of the prescription.

B. -4.00 +2.00 x 090. To transpose, first move the total cylindrical value to the spherical portion. In this example, -2.00 + (-2.00) = -4.00. Once you have the sphere, reverse the symbol of the DC and flip the axis of the original equation.

A. The entrance pupil is defined as the image of the aperture stop formed by all lenses in front of it. The exit pupil is the image of the aperture stop formed by all lenses behind it. An entrance port is the image of the field stop formed by all lenses in front of it. The image of the field stop formed by lenses behind it is known as the exit port.

C. YAG laser is used for LPI, PCO, and SLT. Argon and Krypton are used for PRP, ALT, retinal tears, and in rare instances LPIs. Excimer/LASIK laser is used for LASIK and PRK. Helium neon is used to illuminate the retina to view the fundus.

C. When compared to a direct ophthalmoscope, the BIO has a larger field of view, larger depth of focus, SMALLER magnification, and an inverted image.

A,C,D: There are two definitions within the ANSI Z87.1 Safety Standard for Impact: high mass impact and high velocity impact. High mass impact is the drop ball test. This test requires a pointed projectile that weighs 500g dropped from 50 inches. The high velocity impact is assessed using a 0.25 inch diameter steel ball fired at 150 feet per second.

D. The lens will need to be moved 2 mm to place the optical center in front of the patient’s eye. Do not forget that TOTAL decentration is 4 mm but each lens only moves half of that. Easy rule of thumb – patient PD < frame PD = move lens in and vice versa

B. You can increase the seg height by any of the following: narrow the pads, move pads down, increase vertex distance, reduce pantoscopic tilt, or shrink the bridge. Consider the opposite in cases where the seg height appears too high.

C. A barrel distortion is created by a minus lens. A plus power lens creates a pincushion effect. The other two options are distractors.

B. Polycarbonate has the lowest Abbe value of all these options. Abbe value is inversely proportional to the level of chromatic aberrations. In other words, the higher the Abbe value, a lower amount of chromatic aberrations will be induced.

D. Total prism induced is 0.8 BI. First, recall Prentice’s law (Total prism = power of lens x decentration in centimeters). In this example, the answer is 0.8. Next, draw a picture of the lens (two triangles sharing one side). We can see that the patient is looking through the base-in portion of the lens.

D. The image is displaced 9 cm to the left. The equation for this example is PrismPower = image displacement (in cm)/distance to screen (in m). This is why the answer is in centimeters. Next, we note that a prism moves images towards its apex. When the base is positioned to the right, the image moves to the left.

B. In order to increase spectacle magnification of a plus lens, you can increase vertex distance, thickness, and/or base curve. To decrease SM of a plus lens, increase the index of refraction. Trends for minus lens are slightly different due to the shape of the lens. To increase SM, increase thickness or base curve. To decrease SM, increase vertex distance or index of refraction.